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Masa 2 moli Al = 2*27g = 54g
Masa 2 moli AlCl₃ = 2*(27g + 3*35,5g) = 267g
Zgodnie z reakcją:
z 54g Al---------uzyskano-------267g AlCl₃
to z 2,7g Al-----uzyskamy------Xg AlCl₃
X= 2,7g * 267g : 54g = 13,35g AlCl₃